Let overrightarrow{a}=hat{i}+alpha hat{j}+bet | vedclass

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(A) Given $\vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k}$,so $|\vec{a}|^2 = 1 + \alpha^2 + \beta^2$.Given $|\vec{b}|^2 = 6$,so $|\vec{b}| = \sqrt

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(A) Given $\vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k}$,so $|\vec{a}|^2 = 1 + \alpha^2 + \beta^2$.Given $|\vec{b}|^2 = 6$,so $|\vec{b}| = \sqrt{6}$.The angle between $\vec{a}$ and $\vec{b}$ is $	heta = \frac{\pi}{4}$.Using the dot product formula: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 	heta = 3\sqrt{2}$.Substituting the values: $|\vec{a}| \cdot \sqrt{6} \cdot \cos(\frac{\pi}{4}) = 3\sqrt{2}$.$|\vec{a}| \cdot \sqrt{6} \cdot \frac{1}{\sqrt{2}} = 3\sqrt{2} \implies |\vec{a}| \cdot \sqrt{3} = 3\sqrt{2} \implies |\vec{a}| = \sqrt{6}$.Thus,$|\vec{a}|^2 = 6$,which implies $1 + \alpha^2 + \beta^2 = 6$,so $\alpha^2 + \beta^2 = 5$.Now,calculate $|\vec{a} 	imes \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 	heta$.$|\vec{a} 	imes \vec{b}|^2 = (6)(6) \sin^2(\frac{\pi}{4}) = 36 \cdot (\frac{1}{\sqrt{2}})^2 = 36 \cdot \frac{1}{2} = 18$.Finally,$(\alpha^2 + \beta^2) |\vec{a} 	imes \vec{b}|^2 = (5)(18) = 90$.

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